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Part D find the average rate of change of F on the interval from X is greater than or equal negative four is less than or equal to X which is less than or equal to three so really the interval that they've depicted right over here and then they say there's no point C between in that interval for which F prime of C is equal to the average rate of change explain why the statement does not contradict the mean value theorem fascinating all right let's do this first part first the average rate of change of F on the interval sounds like a very fancy thing but the average rate of change on the interval is really just the slope of the line that connects the endpoints of the interval so these are this right over here those are the endpoints and let's figure out the slope of that line so going we going from this point to that point our change in X we just in a color that you're likely to see our change in x over here our change in X change in X is equal to seven and you could get that by taking three minus negative four you can literally discount one two three four five six seven that's our change in X and our change in Y when we when we run over seven when we go seven to the right our change in Y our change in Y is equal to negative two we went from negative 1 to negative three is equal to negative two so slope slope which is change in Y over change in X rise over run is equal to 7 over negative 2 or oh actually let me know I the other way around change in Y is negative two over seven so negative two sevens and that's all it is now you could you could kind of think of it in kind of a more fancy sense but you're gonna get the exact same answer if you said oh well look you know the average their average rate of change of F on the interval well the rate of change of f is f prime F prime of X that is the rate of change of F at any point X and so if you want to find the average value of this over the interval you would integrate it from our starting point from negative 4 to 3 DX and then you would divide it by your change in X so then you would divide it by you would have it over 1 over 7 over 1 over 7 but then this this part right over here is just going to be the same thing as F of 3 minus F of 4 and then this over here you have a 7 in the denominator so this is really just your change in X this is your change in X which is which it was by definition or how we how we actually set up this average right over here and this over here is really just your change in Y so it really is just the slope between the endpoints so we did the first part our average rate of change of F on the interval is negative 2/7 and then let's think about the second part they say there's no point C in that interval for which F prime of C is equal to the average rate of change explain why the statement does not contradict the mean value theorem so the mean value theorem just is a little bit of a review as a little bit of review it says that if we have some type of an interval if you have an interval so let me draw some axes right over here if you have an interval let me draw an interval like this and over that interval you have a differentiable function you have a differentiable function so maybe my function looks like this it says at least there's at least one point C on that interval where the derivative at that point C is equal to the average rate of change so the way I've drawn it right over here the average rate of change of this function I'll do it in magenta is this right over here and the mean value theorem says is that if this is differentiable there's at least one point C in this interval where I have the same slope where the tangent line has the same slope where the derivative is the same as the slope as that average slope and you can see here it was probably right over here you have one of those points and actually we probably have multiple of them we probably have another point right over here that's like that and then another point there that's like that and if you think about it it's kind of it's kind of intuitive that at some point you know here where we obviously have a larger slope and over here we have a smaller slope and since our since it's differentiable our derivative is continuous so at some point at some point the slope has to get to exactly what the average slope is now let's think about our little conundrum with this question right over here why is there not a Point C for which F prime of C is equal to the average rate of change and you can even verify that for yourself because from negative 4 to 0 our slope is positive we have a positive slope here and then the slope just jumps down to negative 2 it just jumps down to negative 2 which is a much more negative slope than this so it never goes to negative 2/7 and the reason is is that this is not a differentiable function at X is equal to 0 it is not differentiable at x equals 0 our slope jumps here and because it's not differentiable the mean value theorem doesn't apply now you can imagine if this was differentiable if this did have a continuous derivative then you would find a point so if this looks something like this instead I'll continued over here if it looks something like this instead where it was it had a continuous derivative then there would be a point where the slope was the same as the average slope maybe it would have been right over there so it's really because it's really because f is not differentiable over the entire interval it does not have a continuous derivative the derivative jumps from a from a positive value approaching 0 here and it just jumps straight down to negative 2 right over here it doesn't go continuously through all the values in between
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